Divide the following complex numbers. $ \dfrac{-20-8i}{-2+5i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-2-5i}$ $ \dfrac{-20-8i}{-2+5i} = \dfrac{-20-8i}{-2+5i} \cdot \dfrac{{-2-5i}}{{-2-5i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-20-8i) \cdot (-2-5i)} {(-2+5i) \cdot (-2-5i)} = \dfrac{(-20-8i) \cdot (-2-5i)} {(-2)^2 - (5i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-20-8i) \cdot (-2-5i)} {(-2)^2 - (5i)^2} = $ $ \dfrac{(-20-8i) \cdot (-2-5i)} {4 + 25} = $ $ \dfrac{(-20-8i) \cdot (-2-5i)} {29} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-20-8i}) \cdot ({-2-5i})} {29} = $ $ \dfrac{{-20} \cdot {(-2)} + {-8} \cdot {(-2) i} + {-20} \cdot {-5 i} + {-8} \cdot {-5 i^2}} {29} $ Evaluate each product of two numbers. $ \dfrac{40 + 16i + 100i + 40 i^2} {29} $ Finally, simplify the fraction. $ \dfrac{40 + 16i + 100i - 40} {29} = \dfrac{0 + 116i} {29} = 4i $